$#nobody$

`x^3 + 3x + 2 = (x + 2)sqrt[x^3 + 2x + 1]`(ĐK bạn tự tìm nhé..)

`<=> x^3 + 2x + (x + 2) = (x + 2)sqrt[x^3 + 2x + 1]`

`<=> x^3 + 2x = (x + 2)sqrt[x^3 + 2x + 1] - (x + 2)`

`<=> x^3 + 2x = (x + 2)(sqrt[x^3 + 2x + 1] - 1)`

`<=> x^3 + 2x = (x + 2) . (x^3 + 2x + 1 - 1)/(sqrt[x^3 + 2x + 1] + 1)`

`<=> x^3 + 2x = (x + 2) . (x^3 + 2x)/(sqrt[x^3 + 2x + 1] + 1)`

`<=> (x^3 + 2x) - (x + 2) . (x^3 + 2x)/(sqrt[x^3 + 2x + 1] + 1) = 0`

`<=> (x^3 + 2x).[1 - (x + 2)/(sqrt[x^3 + 2x + 1] + 1)] = 0`

`<=> x(x^2 + 2).[1 - (x + 2)/(sqrt[x^3 + 2x + 1] + 1)] = 0`

`TH1 : x = 0(tm)`

`TH2 : x^2 + 2 = 0 <=> x^2 = - 2`(vô lý)

`TH3 : 1 - (x + 2)/(sqrt[x^3 + 2x + 1] + 1) = 0`

`<=> (x + 2)/(sqrt[x^3 + 2x + 1] + 1) = 1`

`<=> x + 2 = sqrt[x^3 + 2x + 1] + 1`

`<=> x + 1 = sqrt[x^3 + 2x + 1]`

`<=> x^2 + 2x + 1 = x^3 + 2x + 1`

`<=> x^3 - x^2 = 0`

`<=> x^2 (x - 1) = 0`

`<=> x^2 = 0` hoặc `x - 1 = 0`

`<=> x = 0(tm)` hoặc `x = 1(tm)`

Vậy : PT đã cho có `2` nghiệm : `x_1 = 0 ; x_2 = 1`